# How do you find the derivative of  f (x) = ln(1 - sin x)?

Mar 18, 2016

$f ' \left(x\right) = \cos \frac{x}{\sin x - 1}$

#### Explanation:

Differentiate using the $\textcolor{b l u e}{\text{ chain rule }}$

$\frac{d}{\mathrm{dx}} \left[f \left(g \left(x\right)\right)\right] = f ' \left(g \left(x\right)\right) . g ' \left(x\right)$

and $\frac{d}{\mathrm{dx}} \left(\ln x\right) = \frac{1}{x}$

$\Rightarrow f ' \left(x\right) = \frac{1}{1 - \sin x} . \frac{d}{\mathrm{dx}} \left(1 - \sin x\right) = - \cos \frac{x}{1 - \sin x} = \cos \frac{x}{\sin x - 1}$